# Using identities prove that x^3+y^3+z^3 > or equal to 3xyz, where x,y, z are positive real numbers?

Consider x^2y^2-z^2. Rewrite x^2y^2-z^2 as left(xy ight)^2-z^2. The difference of squares can be factored using the rule: a^2-b^2=left(a-b ight)left(a+b ight).

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You are correct, this is a Thue equation. "bivariate" just means "having two variables", và "form" refers khổng lồ the polynomial being homogeneous, which it is since every term has degree 3. ...
How vì you multiply displaystyleleft(-x^3y^4 ight)left(2x^2y^5 ight) ?
displaystyle=-2x^5y^9 Explanation: displaystyleleft(-x^3y^4 ight)left(2x^2y^5 ight) As per the property of exponents : displaystyleleft(a^m imesa^n=a^left(m+n ight) ight. ...
2x3y2-6xy3 Final result : 2xy2 • (x2 - 3y) Step by step solution : Step 1 :Equation at the kết thúc of step 1 : ((2 • (x3)) • (y2)) - (2•3xy3) Step 2 :Equation at the end of step 2 : (2x3 • y2) - ...
8x3y2-4xy3 Final result : 4xy2 • (2x2 - y) Step by step solution : Step 1 :Equation at the over of step 1 : ((8 • (x3)) • (y2)) - 22xy3 Step 2 :Equation at the kết thúc of step 2 : (23x3 • y2) - ...

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What are the critical points of displaystylefleft(x,y ight)=x^3+y^3-xy ?
displaystyleleft(0,0 ight)anddisplaystyleleft(frac13,frac13 ight) Explanation: displaystylef_x=3x^2-ydisplaystylef_y=3y^2-x ...
Use implicit differentiation lớn find the largest y-value in the loop of the Folium of Descartes, which is given by : x^3 + y^3 - 3xy = 0.
https://math.stackexchange.com/questions/1021017/use-implicit-differentiation-to-find-the-largest-y-value-in-the-loop-of-the-foli
First, we bởi vì the implicit derivative khổng lồ simplify our equation. Because maxima/minima occur when f'(x)=0, we take the implicit derivative và set it equal khổng lồ 0. We should end up with a few ...

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Consider x^2y^2-z^2. Rewrite x^2y^2-z^2 as left(xy ight)^2-z^2. The difference of squares can be factored using the rule: a^2-b^2=left(a-b ight)left(a+b ight).
left< eginarray l l 2 và 3 \ 5 và 4 endarray ight> left< eginarray l l l 2 & 0 & 3 \ -1 & 1 và 5 endarray ight>

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